Let a,b,c be positive real numbers in H.P.
Statement -1: `(a+b)/(2a-b)+(c+b)/(2c-b)ge4`
Statement-2: `(a)/(b)+(b)/(c)+(c)/(a)ge3`

Options

A. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.
B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.
C. Statement -1 is true, Statement -2 is False.
D. Statement -1 is False, Statement -2 is True.

Explanation

To solve the problem step by step, we will analyze both statements given that \( a, b, c \) are positive real numbers in Harmonic Progression (H.P.).

### Step 1: Understanding Harmonic Progression
If \( a, b, c \) are in H.P., then the reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (A.P.). This means:
\[ \frac{1}{b} = \frac{1}{2} \left( \frac{1}{a} + \frac{1}{c} \right) \]
From this, we can express \( b \) in terms of \( a \) and \( c \):
\[ b = \frac{2ac}{a+c} \]

### Step 2: Analyzing Statement 1
We need to prove:
\[ \frac{a+b}{2a-b} + \frac{c+b}{2c-b} \geq 4 \]
Substituting \( b = \frac{2ac}{a+c} \) into the expression:

1. **Calculate \( a + b \)**:
\[ a + b = a + \frac{2ac}{a+c} = \frac{a(a+c) + 2ac}{a+c} = \frac{a^2 + ac + 2ac}{a+c} = \frac{a^2 + 3ac}{a+c} \]

2. **Calculate \( 2a - b \)**:
\[ 2a - b = 2a - \frac{2ac}{a+c} = \frac{2a(a+c) - 2ac}{a+c} = \frac{2a^2 + 2ac - 2ac}{a+c} = \frac{2a^2}{a+c} \]

3. **Calculate \( c + b \)**:
\[ c + b = c + \frac{2ac}{a+c} = \frac{c(a+c) + 2ac}{a+c} = \frac{ac + c^2 + 2ac}{a+c} = \frac{c^2 + 3ac}{a+c} \]

4. **Calculate \( 2c - b \)**:
\[ 2c - b = 2c - \frac{2ac}{a+c} = \frac{2c(a+c) - 2ac}{a+c} = \frac{2ac + 2c^2 - 2ac}{a+c} = \frac{2c^2}{a+c} \]

5. **Substituting back into the inequality**:
\[ \frac{a+b}{2a-b} + \frac{c+b}{2c-b} = \frac{\frac{a^2 + 3ac}{a+c}}{\frac{2a^2}{a+c}} + \frac{\frac{c^2 + 3ac}{a+c}}{\frac{2c^2}{a+c}} = \frac{a^2 + 3ac}{2a^2} + \frac{c^2 + 3ac}{2c^2} \]

6. **Simplifying the expression**:
\[ = \frac{1}{2} \left( \frac{a^2 + 3ac}{a^2} + \frac{c^2 + 3ac}{c^2} \right) = \frac{1}{2} \left( 1 + \frac{3c}{a} + 1 + \frac{3a}{c} \right) = 1 + \frac{3}{2} \left( \frac{c}{a} + \frac{a}{c} \right) \]

7. **Applying AM-GM inequality**:
\[ \frac{c}{a} + \frac{a}{c} \geq 2 \implies 1 + \frac{3}{2} \cdot 2 \geq 4 \]

Thus, Statement 1 is proven:
\[ \frac{a+b}{2a-b} + \frac{c+b}{2c-b} \geq 4 \]

### Step 3: Analyzing Statement 2
We need to prove:
\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3 \]
Using the AM-GM inequality:
\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3 \sqrt[3]{\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3 \]
Thus, Statement 2 is also proven.

### Conclusion
Both statements are true, but Statement 2 does not explain Statement 1.